Which Vectors Can Legitimately be Broken into Components?

[jon4444]

I ran across the explanation below of a car on a banked turn (no friction) on a High School teacher's website. I'm assuming this is incorrect, because how can the normal force in this situation equal mg/cos when an object on an inclined plane has a normal force equal to mg * cos? The high school teacher takes the normal force and breaks it into components and in the inclined plane we take the force of gravity and break it into components.
If someone could explain the discrepancy / fallacy here, I shall give them my first born:

A free-body diagram for the car on the banked turn is shown at left. The banking angle between the road and the horizontal is

(theta). The normal force, N, has been resolved into horizontal and vertical components (the blue vectors).

In the vertical direction there is no acceleration, and:

so:

 

[kuruman]

The normal force, being a contact force, is not cast in stone but adjusts itself to provide the observed acceleration. The acceleration in the two cases that you mentioned are different, therefore the normal force is expected to be different.

Please keep your first-born, I already have one.

 

[jbriggs444]

If I understand the concern, it is between breaking the normal force up into components in the one case and the gravitational force into components in the other.

The short answer is that it does not matter. You can choose any coordinate system you like. The physics will still work out in the end. However, a good choice of coordinate system can simplify the algebra.

Consider the typical problem of a block sliding down an inclined plane. You have normal force (perpendicular to the slope), friction (parallel to the slope), acceleration (parallel to the slope) and gravity (at an angle to the slope). If you have to choose a coordinate system, you want one where most of your unknowns align with the coordinate system. There is a clear choice: Align your coordinate system with the slope. Put the X axis on the slope and put your Y axis at right angles.

But that is not the problem at hand. We are considering a car on a banked track. We have normal force (perpendicular to the slope), gravity (at an angle to the slope), centripetal acceleration (at an angle to the slope) and friction (arbitrarily set to zero). This time there is a different choice. Put the X axis on the level ground and aim the Y axis vertically at the sky.

The claim is that the algebra still works out either way. Let us demonstrate that. Let us work the banked track scenario first with a coordinate system aligned with the track and then with a coordinate system aligned with the ground.

Analysis in the track coordinate system:

Gravity has a non-zero x component ($mg \sin \theta$) and a non-zero y component ($-mg \cos \theta$). The normal force has a zero x component and a non-zero y component ($N$). Centripetal acceleration $\vec{a}$ has a non-zero x component ($a_x = a \cos \theta$) and a non-zero y component ($a_y = a \sin \theta$). Friction is zero.

Start with Newton's second law:$$\vec{F}=m\vec{a}$$
Adding up the forces in the x direction $$ma_x = mg \sin \theta$$
Dividing out the $m$:$$a_x = g \sin \theta$$
Solving for the magnitude of the centripetal acceleration, we get $$a = \frac{g \sin \theta}{\cos \theta}$$
Which then gives us the y component of the centripetal acceleration, $$a_y = \frac{g \sin^2 \theta}{\cos \theta}$$
Adding up the forces in the y direction, we have $$ma_y = N - mg \cos \theta$$
Plugging in the previously determined formula for $a_y$ we obtain $$m\frac{g \sin^2 \theta}{\cos \theta} = N - mg \cos \theta$$
Multiplying through by $\cos \theta$ we get$$mg \sin^2 \theta = N \cos \theta - mg \cos^2 \theta$$
Adding $mg \cos^2 \theta$ to both sides:$$mg \sin^2 \theta + mg \cos^2 \theta = N \cos \theta$$
But $sin^2 \theta + cos^2 \theta = 1$. So we get$$mg = N \cos \theta$$
Or:$$N = \frac{mg}{\cos \theta}$$Analysis in the ground coordinate system:

Gravity has a zero x component and a non-zero y component ($-mg$). The normal force has a non-zero x component $N \sin \theta$ and a non-zero y component $N \cos \theta$. The centripetal acceleration has a non-zero x component and a zero y component. Friction is zero.

Start with Newton's second law:$$\vec{F}=m\vec{a}$$
Adding up the forces in the y direction:$$-mg + N \cos \theta = 0$$
We can immediately obtain $$N \cos \theta = mg$$
Or:$$N = \frac{mg}{\cos \theta}$$

 

[jon4444]

The normal force, being a contact force, is not cast in stone but adjusts itself to provide the observed acceleration. The acceleration in the two cases that you mentioned are different, therefore the normal force is expected to be different.

Please keep your first-born, I already have one.

My confusion is more on the conceptual level. The only difference between the two cases is the centripetal acceleration in the banked turn case. This leads to a contact force in the x direction. So, it shouldn't change the Y component of the normal force, which should have a magnitude of mg. But, if that magnitude of the normal force in the y direction is N*cos, then that value will increase as the normal force increases, which occurs when the velocity increases. But how can that be if it has to stay equal to mg, which is a constant?

 

[jbriggs444]

My confusion is more on the conceptual level. The only difference between the two cases is the centripetal acceleration in the banked turn case. This leads to a contact force in the x direction. So, it shouldn't change the Y component of the normal force, which should have a magnitude of mg. But, if that magnitude of the normal force in the y direction is N*cos, then that value will increase as the normal force increases, which occurs when the velocity increases. But how can that be if it has to stay equal to mg, which is a constant?

If there is no friction then any increase in normal force must be accompanied by an increase in the bank angle. You cannot just change the one without also adjusting the other. Otherwise the car will not stay on the track.

One could use a bowl-shaped track and have the car simply ride higher on the track, thereby semi-automatically adjusting the bank angle.

 

[kuruman]

My confusion is more on the conceptual level. The only difference between the two cases is the centripetal acceleration in the banked turn case. This leads to a contact force in the x direction. So, it shouldn't change the Y component of the normal force, which should have a magnitude of mg. But, if that magnitude of the normal force in the y direction is N*cos, then that value will increase as the normal force increases, which occurs when the velocity increases. But how can that be if it has to stay equal to mg, which is a constant?

Are you comparing the frictionless banked curve with the block at rest on an incline? If so, then in both cases the incline exerts a force equal to mg in the vertical direction. The difference is that in the banked curve case, the incline exerts and additional horizontal force mv²/r towards the center of the circular path whereas in the other case the incline exerts zero horizontal force. I am repeating what @jbriggs444 said, only differently.

 

[jon4444]

Are you comparing the frictionless banked curve with the block at rest on an incline? If so, then in both cases the incline exerts a force equal to mg in the vertical direction. The difference is that in the banked curve case, the incline exerts and additional horizontal force mv²/r towards the center of the circular path whereas in the other case the incline exerts zero horizontal force. I am repeating what @jbriggs444 said, only differently.

Yes, but then why in the banked curve case does the vertical component of the normal force equal N/(cos theta), which should increase with increasing velocity, but at the same time equal mg, which is a constant?

 

[jbriggs444]

Yes, but then why in the banked curve case does the vertical component of the normal force equal N/(cos theta), which should increase with increasing velocity, but at the same time equal mg, which is a constant?

Asked and answered: If you increase velocity without changing the bank angle and without downward friction, you fly off the track.

 

[jon4444]

Asked and answered: If you increase velocity without changing the bank angle and without downward friction, you fly off the track.

Is that because the vertical component of the Normal force is now greater than mg?

 

[jbriggs444]

Is that because the vertical component of the Normal force is now greater than mg?

Yes.

 

[jon4444]

Yes.

Why?
This gets to the nub of my confusion. Why does increasing velocity increase the vertical component of the Normal force. I would have thought it just increases the x component of the normal force since the increased "contact" force must point to the center of track. If that increased component is in the x direction only (since it arises from need for increased centripetal force), why does the y component increase?

 

[jbriggs444]

Why?
This gets to the nub of my confusion. Why does increasing velocity increase the vertical component of the Normal force. I would have thought it just increases the x component of the normal force since the increased "contact" force must point to the center of track. If that increased component is in the x direction only (since it arises from need for increased centripetal force), why does the y component increase?

The normal force is the normal force. That is to say it is a force perpendicular to the surface. That means that its direction is fixed. One cannot increase its x component without increasing its y component.

 

[jon4444]

The normal force is the normal force. That is to say it is a force perpendicular to the surface. That means that its direction is fixed. One cannot increase its x component without increasing its y component.

Sorry, that makes no sense to me. It's direction is fixed, but it increases as velocity increases. That increase is in a purely x direction (correct?) therefore it's magnitude in the y direction would stay the same.
Unless what you'r saying is that as velocity increases, the normal force increases in a direction perpendicular to the track (not just in the direction of centripetal force)--this is sort of what I keep asking, but I can't get clarification on whether this is the case.

 

[jbriggs444]

Edit: @jon4444 I think I now better understand what you are trying to say:

Unless what you'r saying is that as velocity increases, the normal force increases in a direction perpendicular to the track (not just in the direction of centripetal force)--this is sort of what I keep asking, but I can't get clarification on whether this is the case.

Yes. The normal force is, by definition, perpendicular to the track. Any increase in the x direction as required to match the centripetal acceleration must be accompanied by a corresponding increase in the y direction. i.e. the car flies up and out of the track.

Do you know what "normal force" means? And do you know how to keep direction of a vector constant while changing its x component and holding the y component fixed?

 

[A.T.]

Unless what you'r saying is that as velocity increases, the normal force increases in a direction perpendicular to the track

The normal force is perpendicular to the track per definition.

 

[jon4444]

What's been confusing me is that someone definitely stated that the 'Y" component of the normal force is always equal to mg. This is not true.
The conceptual framework I've been looking for might sound like this: at speeds above "ideal", the normal force increases and y component is above mg. If the car is in equilibrium, it'd be because static friction acts down the slope of the track adding an additional downward force to gravity to balance out the increase in normal force in the up y direction.
Kosher?

 

[kuruman]

What's been confusing me is that someone definitely stated that the 'Y" component of the normal force is always equal to mg. This is not true.

I agree. If by "Y component" you mean the vertical component of the normal force, then it is equal to mg only when the acceleration in the vertical direction is zero, i.e. when the car goes around the banked curve in a horizontal circle. When this Y component is not equal to mg, then the car will not be able to go around in the horizontal circle as @jbriggs444 has repeatedly indicated. Consider the following instructive problem.

A puck of mass $m$ is placed on a frictionless plane inclined at angle $\alpha$ with respect to the horizontal. Coordinate axes are drawn on the incline such that the x-axis is horizontal and the positive y-axis points straight uphill. At some instant, the puck is moving with velocity $\vec v=v_{x}\hat x+v_{y}\hat y$. Find the magnitude of the normal force exerted by the plane on the puck at that instant.

Edit: The question is more complete if it is given that the puck starts from the origin with velocity $\vec {v}_0=v_{0x}\hat x+v_{0y}\hat y$.

 

[A.T.]

The conceptual framework I've been looking for might sound like this: at speeds above "ideal", the normal force increases and y component is above mg. If the car is in equilibrium, it'd be because static friction acts down the slope of the track adding an additional downward force to gravity to balance out the increase in normal force in the up y direction.

I don't think it is conceptually sensible to think of friction as balancing an increase in normal force, because they are orthogonal.

I think it's better to say that the total contact force (which has normal and friction components) adjusts to balance gravity, and provide the needed centripetal acceleration.